Can't infer constrained types

[falsandtru]

@ahejlsberg You mean this?

Note that is not possible for a conditional type to recursively reference itself, as might be desired in the Unpacked case above. We're still considering ways in which to implement this.

TypeScript Version: master

Search Terms:

Code

type A<T, U extends A<any, any>> = [T, U];
type B<S> = S extends A<infer T, infer U> ? [T, U] : never;

Expected behavior:

pass

Actual behavior:

$ node built/local/tsc.js --noEmit index.ts
index.ts(2,34): error TS2344: Type 'U' does not satisfy the constraint '[any, an
y]'.

Related Issues:

[falsandtru]

No, probably this:

The definitely assignable relation is the same as the regular assignable relation, except that type variable constraints are not considered.

We may at a later date support infer declarations in regular type parameter constraints, however it is not trivial to implement. Meanwhile, you can get much the same effect with the existing constraint mechanism as I explain in the introduction to the PR.

#21496 (comment)

type A<T, U extends void> = [T, U];
type B<S> = S extends A<infer T, infer U> ? [T, U] : never;
type C<S, U extends void> = S extends A<infer T, U> ? [T, U] : never;

$ node built/local/tsc.js --noEmit index.ts
index.ts(2,34): error TS2344: Type 'U' does not satisfy the constraint 'void'.